Question

Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? f(x) = 4x2 − 3x + 2, [0, 2]

Answer 1

Answer:

Yes , it satisfies the hypothesis and  we can conclude that c = 1 is an element of [0,2]

c = 1 ∈ [0,2]

Step-by-step explanation:

Given that:

$f(x) = 4x^2 -3x + 2, [0, 2]$

which is read as the function of x = 4x² - 3x + 2 along the interval [0,2]

Differentiating the function with respect to x is;

f(x) = 8x - 3

Using the Mean value theorem to see if the function satisfies it, we have:

$f'c = \dfrac{f(b)-f(a)}{b-a}$

$8c -3 = \dfrac{f(2)-f(0)}{2-0}$

since the polynomial function is differentiated in [0,2]

$8c -3 = \dfrac{(4(2)^2-3(2)+2)-(4(0)^2-3(0)+2)}{2-0}$

$8c -3 = \dfrac{(4(4)-3(2)+2)-(4(0)-3(0)+2)}{2-0}$

$8c -3 = \dfrac{(16-6+2)-(0-0+2)}{2-0}$

$8c -3 = \dfrac{(12)-(2)}{2}$

$8c -3 = \dfrac{10}{2}$

8c -3  = 5

8c = 5+3

8c = 8

c = 8/8

c = 1

Therefore, we can conclude that c = 1 is an element of [0,2]

c = 1 ∈ [0,2]